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Find the square root of the algebraic expression which is then average of the following expressions

$x^{2} + \frac{1}{x^{2}}, - 2 (x - \frac{1}{x}) and - 1$

Subtopics: Degree of polynomials

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\frac{1}{\sqrt{3}} |x + \frac{1}{x} + 1|

\frac{1}{\sqrt{3}} |x - \frac{1}{x} + 1|

\frac{2}{\sqrt{3}} |x - \frac{1}{x} - 1|

\frac{1}{\sqrt{3}} |2 x + \frac{1}{x} + 1|

answer is B.

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Detailed Solution

Average = \frac{sum of observation}{Total number of observation}

Average = \frac{x^{2} + \frac{1}{x^{2}} - 2 (x - \frac{1}{x}) - 1}{3}

Average = \frac{x^{2} + \frac{1}{x^{2}} - 2 (x - \frac{1}{x}) - 1 + 2 - 2}{3}

Average = \frac{x^{2} - 2 + \frac{1}{x^{2}} - 2 (x - \frac{1}{x}) - 1 + 2}{3}

Average = \frac{x^{2} - 2 + \frac{1}{x^{2}} - 2 (x - \frac{1}{x}) + 1}{3}

By using formula

{(a - b)}^{2} = a^{2} - 2 ab + b^{2}

{(x - \frac{1}{x})}^{2} = x^{2} - 2 x (\frac{1}{x}) + {(\frac{1}{x})}^{2}

{(x - \frac{1}{x})}^{2} = x^{2} - 2 + \frac{1}{x^{2}}

Average = \frac{{(x - \frac{1}{x})}^{2} - 2 (x - \frac{1}{x}) + 1}{3}

By using same formula

{(a - b)}^{2} = a^{2} - 2 ab + b^{2}

Average = \frac{{(x - \frac{1}{x} - 1)}^{2}}{3}

\sqrt{Average} = \sqrt{\frac{{(x - \frac{1}{x} - 1)}^{2}}{3}} = \frac{\sqrt{{(x - \frac{1}{x} - 1)}^{2}}}{\sqrt{3}} = \frac{|x - \frac{1}{x} - 1|}{\sqrt{3}}

The square root of any real number is always positive.

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